2 SPECTRUM OF AN ELEMENT OF A NORMED ALGEBRA 311 (15.2.3) Let P(X) = a0 4- oqX H ----- h anXn be a nonconstant polynomial with complex coefficients, with ocn ^ 0. For each x e A, put P(x) = a0 e 4- #1* + + anxn e A. For each complex number /^, we can write P(X) n in the form where the Aj are complex numbers (9.11.2). Since the subalgebra of A genera- ted by e and x is commutative, we have P(x) - jw? = an(x - V) "(*- AB e). Hence P(x) \JLQ is invertible in A if and only if all the x ^e are invertible in A; consequently \JL e SpA(P(Xŧ if and only if lj e SpA(x) for at least one value of y, and since JLI = P(/lj) for each j, this condition is equivalent to \i e P(SpA(x)). Therefore (15.2.3.1) SpA(P(x)) = P(SpA(x)). Suppose now that 0 ^ P(SpA(;c)), so that P(*) is invertible in A. Let Q(X) be another polynomial with complex coefficients, and put R(X) = Q(X)/P(X). Then we may form the element Q(x)(P(;c))~1 = (P(x))~lQ(x) of A, which we denote by R(x). Then we have (15.2.3.2) SpA(R(*)) = R(SpA(*)). For, replacing R by R ^u for some /i e C if necessary, we have to show that R(x) is invertible if and only if 0 ^ R(SpA(x)); or equivaiently that Q(x) is invertible if and only if 0 Ģ Q(SpA(X)). But this follows from (15.12.3) if Q is not a constant, and it is obvious if Q is a constant. (1 5.2.4) Let A be a Banach algebra with unit element e^Q. (i) The multiplicative group G of invertible elements of A is open in A, and contains the ball \\x e\\ < 1. The topology induced on G by that of A. is compatible with the group structure of G. (ii) For each x e A, the set Rx = C SpA(x) of regular values, for x is open in C, and the mapping C i > (x (e)~ * of Rx into A is analytic. (iii) The set SpA(x) is compact and nonempty, and is contained in the ball ici ^ n*ii- Every C e C such that x Ģe is a (left or right) zero-divisor in A is a